<!DOCTYPE html>
<html lang="en-US">
<!--********************************************-->
<!--*       Generated from PreTeXt source      *-->
<!--*                                          *-->
<!--*         https://pretextbook.org          *-->
<!--*                                          *-->
<!--********************************************-->
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<meta name="robots" content="noindex, nofollow">
</head>
<body class="ignore-math">
<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>(ii) Consider</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_17.html ./knowl/eq3_18.html ./knowl/eq3_21.html ./knowl/eq3_19.html ./knowl/eq3_20.html">
\begin{equation}
y^{\prime \prime}+b y^{\prime}+c y=e^{\alpha x} P_n(x).\tag{3.6.11}
\end{equation}
</div>
<p class="continuation">Introduce</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_17.html ./knowl/eq3_18.html ./knowl/eq3_21.html ./knowl/eq3_19.html ./knowl/eq3_20.html">
\begin{equation*}
y=u(x) e^{\alpha x}.
\end{equation*}
</div>
<p class="continuation">Substituting it into the ODE, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_17.html ./knowl/eq3_18.html ./knowl/eq3_21.html ./knowl/eq3_19.html ./knowl/eq3_20.html">
\begin{equation*}
\begin{aligned}
&amp;\alpha^2 e^{\alpha x} u+ 2 \alpha u^{\prime} e^{\alpha x}+u^{\prime \prime} e^{\alpha x}+b (u^{\prime} +\alpha u )e^{\alpha x}+c u e^{\alpha x}=e^{\alpha x} P_n(x)\\
&amp;\to u^{\prime \prime}+(2 \alpha+b)u^{\prime}+(\alpha^2+b \alpha+c) u=P_n(x).
\end{aligned}
\end{equation*}
</div>
<p class="continuation">It is the form of (<a href="" class="xref" data-knowl="./knowl/eq3_17.html" title="Equation 3.6.6">(3.6.6)</a>) in part (i).<dfn class="terminology">If</dfn> <span class="process-math">\(\alpha^2+b \alpha+c \neq 0\text{,}\)</span> we can assume that the solution <span class="process-math">\(u\)</span> have the form (<a href="" class="xref" data-knowl="./knowl/eq3_18.html" title="Equation 3.6.7">(3.6.7)</a>). In other words, we can assume the particular solution to (<a href="" class="xref" data-knowl="./knowl/eq3_21.html" title="Equation 3.6.11">(3.6.11)</a>) is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_17.html ./knowl/eq3_18.html ./knowl/eq3_21.html ./knowl/eq3_19.html ./knowl/eq3_20.html">
\begin{equation*}
Y=e^{\alpha x} (A_0 x^n+a_1 x^{n-1}+\cdots+A_{n-1}x+A_n).
\end{equation*}
</div>
<p class="continuation"><dfn class="terminology">If</dfn> <span class="process-math">\(\alpha^2+b \alpha+c=0\)</span> and <span class="process-math">\(2 \alpha+b \neq 0\text{,}\)</span> ( i.e., <span class="process-math">\(\alpha\)</span> is a root (not repeated root) of the characteristic equation <span class="process-math">\(r^2+b r+c=0\)</span>), then <span class="process-math">\(u\)</span> has the form (<a href="" class="xref" data-knowl="./knowl/eq3_19.html" title="Equation 3.6.9">(3.6.9)</a>). In other words, we can assume</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_17.html ./knowl/eq3_18.html ./knowl/eq3_21.html ./knowl/eq3_19.html ./knowl/eq3_20.html">
\begin{equation*}
Y=x e^{\alpha x} (A_0 x^n+a_1 x^{n-1}+\cdots+A_{n-1}x+A_n).
\end{equation*}
</div>
<p class="continuation"><dfn class="terminology">If</dfn> both <span class="process-math">\(\alpha^2+b \alpha+c=0\)</span> and <span class="process-math">\(2 \alpha+b=0\)</span> which implies</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_17.html ./knowl/eq3_18.html ./knowl/eq3_21.html ./knowl/eq3_19.html ./knowl/eq3_20.html">
\begin{equation*}
\begin{aligned}
&amp;\alpha=-\frac{b}{2}, \quad \left(-\frac{b}{2}\right)^2+b \left(-\frac{b}{2}\right)+c=0, \\
&amp;\to -\frac{b^2}{4}+c=0,\\
&amp;\to b^2-4 c=0.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Therefore, <span class="process-math">\(\alpha=-\frac{b}{2}\)</span> is a repeated root of <span class="process-math">\(r^2+b r+c=0\text{.}\)</span> Then <span class="process-math">\(u\)</span> should have the form (<a href="" class="xref" data-knowl="./knowl/eq3_20.html" title="Equation 3.6.10">(3.6.10)</a>). In other words,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_17.html ./knowl/eq3_18.html ./knowl/eq3_21.html ./knowl/eq3_19.html ./knowl/eq3_20.html">
\begin{equation*}
Y=x^2 e^{\alpha x} (A_0 x^n+a_1 x^{n-1}+\cdots+A_{n-1}x+A_n).
\end{equation*}
</div>
<span class="incontext"><a href="sec3_6.html#p-132" class="internal">in-context</a></span>
</body>
</html>
